Practice questions · Topic 4.5

Unit 4 · Practice for the Topic 4.5 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Topic 4.5 summary, part 1: the cell cycle

Video coming soon

Grow, copy, divide: G1, S phase, G2, mitosis, cytokinesis; count centromeres, and the DNA per cell halves only at cytokinesis; one copy and one separation give each daughter cell everything.

Watch next: Topic 4.5 summary, part 2: a slide read as data

Video coming soon

Percent in a stage is the count ÷ total × 100, the mitotic index; time in a stage is its fraction × the given cycle length; tissues are compared with error bars, percent change and pile-ups.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent. Where a question gives the length of the whole cell cycle, use that length. Give times in hours to one decimal place and in minutes to the nearest minute.

Question 1

Four cases of cells dividing are described.

Which of the following is cell division for asexual reproduction?

A. An oat seedling’s first leaf lengthens by 3 cm in two days B. A potato plant grows a whole new plant from a bud on one of its tubers C. Bark grows back over a wound on a tree trunk D. An apple tree makes seeds after bees have pollinated its flowers

Question 2

Between divisions, a stained cell from a rice root shows an evenly grainy nucleus and no separate rods.

Why is the cell's DNA in this form between divisions?

A. Grainy DNA is DNA that has been broken into short pieces ready for copying B. The rods have already been handed on to the daughter cells C. The chromosomes have coiled up tightly so they can be moved D. Each chromosome is spread out as chromatin, open to be read and copied

Question 3

An elephant body cell has 56 chromosomes. In S phase it copies its DNA.

How many chromatids does the cell hold at the end of S phase?

A. 28 B. 56 C. 112 D. 224

Question 4

A cell in a growing fern root has just finished cytokinesis and become two daughter cells.

Which three stages does each daughter cell pass through next, in order?

A. G1, S phase, G2 B. S phase, G1, G2 C. G2, S phase, G1 D. Mitosis, G1, S phase

Question 5

Cells of one newt species hold about 40 pg of DNA in G1.

Which DNA reading shows a newt cell part-way through S phase?

A. 20 pg B. 40 pg C. 60 pg D. 80 pg

Question 6

A cell from a growing lettuce root is watched under a microscope.

Which of the following describes the cell while it is in interphase?

A. Its chromosomes are coiled into rods and lined up in one row across its middle B. Its nuclear envelope has broken down and spindle fibers are reaching its chromosomes C. Its cytoplasm is being divided in two by a cell plate D. Its nucleus looks grainy with no separate chromosomes visible, and it is growing

Question 7

Of the cells on a slide from the tip of a rye root grown at 20 °C, 15% are in mitosis. Of the cells on a slide from a root of the same plant grown at 5 °C, 6% are in mitosis.

What is the percent change in the percent of cells in mitosis, from the 20 °C root to the 5 °C root?

A. −150% B. −60% C. −9% D. +60%

Question 8

Two nuclei have formed in a dividing cell from a frog embryo, and the cytoplasm is now being divided.

Which structure divides the cytoplasm of this cell, and how is it made?

A. A cleavage furrow, made by a ring of protein under the membrane tightening B. A cell plate, made by vesicles fusing at the equator into a new cell wall C. A nuclear envelope, made by the spindle fibers joining up across the middle D. A centromere, made by the sister chromatids joining at the middle of the cell

Question 9

A skin cell at the edge of a healing scrape divides again and again.

Which of the following describes the cell cycle?

A. The division of a cell’s nucleus into two nuclei, each with a full set of chromosomes B. The repeating sequence in which a cell grows, copies its DNA and divides in two C. The movement of the separated chromatids from the equator to the two poles of the cell D. The long stretch between one division and the next, in which the cell rests

Question 10

A body cell of an alligator holds 32 chromosomes.

Which of the following is the cell’s genome?

A. The 16 chromosomes that each daughter cell receives when the cell divides B. The set of proteins that the 32 DNA molecules are wound around C. The single longest of the 32 DNA molecules D. The complete set of 32 chromosomes, which each daughter cell receives in full

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Scientific Investigation · 5 points

An earthworm that has lost the last segments of its tail grows them back over a few weeks. A student asks whether more of the cells in the regrowing tail are dividing than in the rest of the body. Five students each fix and stain one such earthworm, cut a thin slice through the regrowing tail and a second slice through the body halfway along the worm, and record the stage of 200 cells in each slice. The table gives one student’s counts for her regrowing-tail slice. The graph gives the mean percent of cells in mitosis for the five regrowing-tail slices and the five mid-body slices, with error bars that represent ±2SE. Take the whole cell cycle in an earthworm as 24 hours.

Top: the stage of each of the 200 cells the student counted on her slide of the regrowing tail. Bottom: the mean percent of cells in mitosis for the five students' regrowing-tail slides and five mid-body slides; error bars represent ±2SE. Gridlines every 5%.

(a) Identify the dependent variable in this investigation. (1 point)

Hint: Which quantity did each student read off her slides and record, as against the thing the students chose to compare?

A full-credit answer: The dependent variable is the percent of the 200 cells in each slice that were in mitosis, the quantity the students measured.

Check the box for each point your answer earns

the dependent variable is the percent of cells in mitosis (or the stage each of the 200 cells was recorded in).

Accept "the number of cells in mitosis out of 200". Do not award the point for "the regrowing tail and the mid-body" (that is the independent variable) or for "200 cells" (a condition kept the same).

Common slip: Naming the part of the earthworm (regrowing tail or mid-body). That is what the students chose to compare, the independent variable; the dependent variable is what they counted.

(b) Calculate the percent of the cells on the regrowing-tail slide that were in mitosis. (1 point)

Hint: Which four stages in the table together make up mitosis, and what total does the percent equation divide their sum by?

Write down the values in the question:

cells in mitosis = 22 + 9 + 9 + 6 = 46

total cells counted = 200

Write down the equation:

percent in mitosis = \frac{cells in mitosis}{total cells counted} \times 100

Substitute the values into the equation:

percent in mitosis = \frac{cells in mitosis}{total cells counted} \times 100

percent in mitosis = \frac{46}{200} \times 100

percent in mitosis = 0.23 \times 100 = 23 %

A full-credit answer: On the regrowing-tail slide 23% of the cells were in mitosis: 46 of the 200 cells were in prophase, metaphase, anaphase or telophase.

Do not award the point for 11% (prophase alone), 77% (the interphase share), about 30% (46 divided by 154, the interphase count) or 22% (the five-slide mean read off the graph).

(c) Calculate the time, in hours to one decimal place, that a regrowing-tail cell spends in metaphase, taking the whole cycle as 24 hours. (1 point)

Hint: Start from the share of the 200 cells that were in metaphase. What does the question give as the length of the whole cycle, and how does the time-in-stage equation use that length?

Write down the values in the question:

cells in metaphase = 9

total cells counted = 200

cycle length = 24 h

Write down the equations:

percent in metaphase = \frac{cells in metaphase}{total cells counted} \times 100

time in metaphase = \frac{percent in metaphase}{100} \times cycle length

Substitute the values into the equations:

percent in metaphase = \frac{9}{200} \times 100 = 4.5 %

time in metaphase = \frac{4.5}{100} \times 24 h

time in metaphase = 0.045 \times 24 h = 1.08 h

1.08 h \approx 1.1 h (to one decimal place)

1.08 h \times 60 min/h = 64.8 min \approx 65 min

A full-credit answer: A regrowing-tail cell spends 1.1 hours in metaphase (1.08 h to two decimal places), about 65 minutes.

Accept 1.08 h. Do not award the point for 4.5 h (the percent read as hours), 0.045 h (the fraction with no cycle length) or 2.2 h (9 divided by 100 instead of 200).

(d) Support the claim that more of the regrowing tail's cells are dividing than of the mid-body's cells, using evidence from the error bars. (1 point)

Hint: What does the caption say each bar represents, and what does the overlap rule say about two bars of that kind?

A full-credit answer: The caption says the bars represent ±2SE, so the overlap rule applies.
The regrowing tail's bar runs from 19.0% to 25.0% of cells in mitosis.
The mid-body's bar runs from 2.4% to 5.6%.
The bars do not overlap.
So the difference between the two groups is very unlikely to be chance.
More of the regrowing tail's cells were caught in mitosis.
So more of the regrowing tail's cells are dividing.

Check the box for each point your answer earns

the evidence (the ±2SE bar for the regrowing tail runs from 19.0% to 25.0% and the bar for the mid-body from 2.4% to 5.6%, and they do not overlap) AND the reasoning (so the higher percent in mitosis in the regrowing tail is very unlikely to be chance, so more of the regrowing tail's cells were caught dividing).

Accept a comparison that quotes both ranges, says they are apart, and links that to the claim. Do not award the point for comparing the two means alone (22.0% against 4.0%), or for the ranges with no link.

Common slip: Comparing 22.0% with 4.0% and stopping, or quoting the bars with no link. Supporting the claim needs the non-overlap and what it shows.

(e) Evaluate the claim that a greater percent of cells in mitosis shows that every regrowing-tail cell cycles faster than a mid-body cell. (1 point)

Hint: One slide freezes every cell at one instant. Think about what a count of frozen cells can tell you directly, and what about a single cell's cycle it cannot.

A full-credit answer: The claim is not supported.
A count of frozen cells shows how many were caught dividing at one instant.
So a greater percent in mitosis shows that more of the regrowing tail’s cells are dividing, or that mitosis takes longer in the regrowing tail.
The speed of a single cell’s cycle is its cycle length.
A count of frozen cells does not give the cycle length.
So the count cannot show that every regrowing-tail cell cycles faster.

Check the box for each point your answer earns

the judgement (the claim is not supported) AND the ground (a count of frozen cells shows how many were caught dividing, so a greater percent shows that more of the regrowing tail's cells are dividing, or that mitosis takes longer there; it cannot show the speed of any one cell's cycle, which would need the cycle length).

Accept any correct statement of what one frozen slide cannot show: whether any single cell cycles faster, or the length of the cycle. Do not award the point for the judgement alone, or for 'every regrowing-tail cell divides faster' stated as what the data show.

Common slip: Judging the claim without the ground, or accepting it. The count says how many of the sampled cells were dividing; it does not show the speed of any one cell.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

The drawing shows two cells from the root of Haplopappus, a small desert daisy whose body cells hold four chromosomes: two long and two short. The cells were fixed and stained at one instant and are labeled cell A and cell B.

Two cells from the root of Haplopappus, a small desert daisy whose body cells hold four chromosomes (two long and two short), fixed and stained at one instant.

(a) Identify the stage of mitosis shown by cell A. Describe the feature of the drawing that shows it. (1 point)

A full-credit answer: Cell A is in anaphase.
The sister chromatids have separated.
Two groups of four single chromatids are moving apart toward opposite poles.
Their centromeres lead the way, pulled by fibers from the centrosomes.
No nuclear envelope surrounds them.

Check the box for each point your answer earns

anaphase, shown by two groups of single chromatids (V shapes with the centromere leading) moving apart toward opposite poles, pulled by the spindle fibers, with no envelope around them.

Accept "the sister chromatids have separated and are moving to the poles". Do not award the point for metaphase (there is no row at the equator) or for telophase (the fibers are still attached and no envelope is forming).

Common slip: Naming telophase because the chromosomes are in two groups. In telophase the fibers are gone and an envelope forms around each group. Here the fibers are still pulling and the groups are still on the move.

(b) Each G1 cell of this plant holds 5 pg of DNA. Calculate the amount of DNA in cell A. (1 point)

Write down the values in the question:

DNA per G1 cell = 5 pg

copies of every chromosome after S phase = 2

Write down the equation:

DNA per cell after S phase = 2 \times DNA per G1 cell

Substitute the values into the equation:

DNA in cell A = 2 \times 5 pg = 10 pg

A full-credit answer: Cell A holds 10 pg of DNA.
Every chromosome was copied in S phase, so the DNA per cell doubled from 5 pg to 10 pg.
The sister chromatids have separated, but the cell has not yet divided, so all 10 pg are still in cell A.

Do not award the point for 5 pg (the G1 amount: the DNA has been copied and the cell has not yet split in two) or 20 pg (doubling twice).

(c) Explain why cell B is dividing its cytoplasm with a plate across its middle rather than with a groove pinching in. (1 point)

A full-credit answer: Cell B is a plant cell with a rigid cell wall.
A cell wall cannot be pinched in the way an animal cell’s membrane can.
So vesicles gather at the equator and fuse into a cell plate, the structure across the middle of the drawing.
The cell plate grows outward until it reaches the old cell wall.
Then the cell plate becomes the new cell wall between the two daughter cells.

Check the box for each point your answer earns

a plant cell's rigid cell wall cannot be pinched in, so vesicles gather at the equator and fuse into a cell plate that grows outward into a new cell wall between the two daughter cells.

Accept "the cell wall cannot be pinched, so a new cell wall is built between the nuclei". Do not award the point for "plant cells have no ring of protein" with no reference to the cell wall, or for describing the cell plate as the spindle.

Common slip: Saying the cell plate is the spindle left behind. The spindle broke down in telophase. The cell plate is new cell wall, built from vesicles fusing at the equator.

(d) Justify the claim that the two cells formed from cell B carry the same genome as each other. (1 point)

A full-credit answer: Before cell B divided, every one of its four chromosomes was copied once in S phase into two identical sister chromatids.
In anaphase the sister chromatids separated.
The spindle sent one chromatid of every pair to each pole.
So each of the two nuclei in cell B holds one copy of every chromosome.
Therefore the two daughter cells carry the same complete genome, four chromosomes each, identical to each other and to the parent.

Check the box for each point your answer earns

every chromosome was copied once in S phase into two identical sister chromatids, and anaphase sent one chromatid of every pair to each pole, so each daughter cell receives one complete copy of the genome: one of each of the four chromosomes, identical to the other daughter's.

Accept "each nucleus holds one copy of every chromosome because each was copied and the copies were separated". Do not award the point for "the parent shared its four chromosomes out, two to each side".

Common slip: Saying the four chromosomes were shared out two and two. The chromosomes were doubled first, in S phase, so each daughter cell receives all four.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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